A particle with velocity $v(t)=t^2+3$, where $t$ is time in seconds, moves in a straight line. How far does the particle move from $t=2$ to $t=3$ seconds?
The definite integral $ \int_a^b |v(t)| \,dt$ gives the total distance traveled by a particle over the interval $[a,b]$. In this case, $ \int_2^3 |t^2+3| \,dt$ represents the total distance traveled by the particle from $t=2$ to $t=3$ seconds. Since $t^2+3$ is always positive, we can rewrite the integral: $ \int_2^3 t^2+3\,dt$ We can now evaluate the integral: $\begin{aligned} \int_2^3 t^2+3 \,dt&=\Big[\dfrac{t^3}{3}+3t\Big]_2^3\\ \\ \\ &=18-\dfrac{26}{3}\\ \\ &=\dfrac{28}{3}\\ \end{aligned}$ The answer: $\dfrac{28}{3}$ units